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Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because
Options:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because
Solution:
1653 Upvotes
Verified Answer
The correct answer is:
$\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
$\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
$$
\begin{aligned}
& {\left[\mathrm{Mn}^{3+}+e^{-}\right.}\left.\longrightarrow \mathrm{Mn}^{2+}\right] \times 4 \\
& 2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 e^{-} \\
& 4 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 \\
& E_{\text {cell }}^{\circ}= 1.50-1.23=0.27 \mathrm{~V} \\
& E_{\text {cell }}^{\circ} \text { is + ve }
\end{aligned}
$$
\begin{aligned}
& {\left[\mathrm{Mn}^{3+}+e^{-}\right.}\left.\longrightarrow \mathrm{Mn}^{2+}\right] \times 4 \\
& 2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 e^{-} \\
& 4 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 \\
& E_{\text {cell }}^{\circ}= 1.50-1.23=0.27 \mathrm{~V} \\
& E_{\text {cell }}^{\circ} \text { is + ve }
\end{aligned}
$$
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